That this set is necessary is proven by Lekkekerker and Boland in [1]} (note that the word acyclic in the referenced lemma means that the graph has no irreducible cycles of length greater than 3, so every interval intersection graph is acyclic in this sense). It will then be enough to find an interval intersection graph \(\mathbb {G}\)  for each \(e\in A_\ell \)  such that \(\mathbb {G}\)  extends \(\mathrm {III}_\ell \) , \(e\)  is an edge of \(\mathbb {G}\) , and no other \(e^{\prime }\in A_\ell \)  is an edge in \(\mathbb {G}\) .
Let \(e=\lbrace i,j\rbrace \)  where \(i\in \lbrace 1,2\rbrace \)  and consider the graph \(\mathbb {G}\)  that is \(\mathrm {III}_\ell \)  with all of the edges \(\lbrace j,k\rbrace \)  for \(k\ne j\)  added. If \(e=\lbrace 1,3\rbrace \) , no new edges incident to 2 are added and if \(j\ne 3\)  then no elements of \(A_\ell \)  are added except \(e\) . We claim that \(\mathbb {G}\)  is an interval intersection graph.
Since \(j\)  is a neighbor of every element of \(\mathbb {G}\) , we may construct a collection of intervals that have an intersection graph isomorphic to the subgraph of \(\mathbb {G}\)  induced by \(G\setminus \lbrace j\rbrace \)  and then take \(j\)  to be the union of these intervals. The induced subgraph looks like
\(\hbox{\begin{tikzpicture} [circle dotted/.style={line width = 0.3mm, dash pattern= on .05mm off 1mm, line cap = round}]{enclosed/.style={draw, circle, inner sep=0pt, minimum size=.1cm, fill=black}}\end{tikzpicture}\node [enclosed, label=above:$(j+1)$] (1) at (0,0) {};\node [enclosed, label=above:$(j+2)$] (2) at (3*0.5,0) {};\node [enclosed, label=above:$\ell $] (3) at (6*0.5,0) {};\node [enclosed, label=above:$1$] (4) at (9*0.5,0) {};\node [enclosed, label=above:$2$] (5) at (12*0.5,0) {};\node [enclosed, label=above:$(j-1)$] (6) at (15*0.5,0) {};}(1) -- (2);[dashed] (2) -- (3);(3) -- (4);(4) -- (5);[dashed] (5) -- (6);\) 
\(which is the intersection graph of a collection of intervals that barely overlap at the endpoints.\) 
The definition of necessary sets for \(\mathbb {G}\)  in \(S\)  implies that any translation of a necessary set for \(\mathbb {G}\)  by an automorphism is a necessary set (in particular, composition with an automorphism of \(\mathbb {G}\)  is a permutation of the injective graph homomorphisms from \(\mathbb {G}\)  to any other graph).
